Epsilon-proof as x->-infinity currently defeating me

**Grep** · December 23rd 2010, 04:31 PM

I think Epsilon-delta proofs and I should spend some quality time together.

The problem is I need to prove that:

$\lim_{x \to -\infty} a^x = 0$

So, I need to show that, for every $\epsilon > 0$ , there is a $\delta$ such that $|f(x) - L| < \epsilon$ whenever $x < \delta$ .

So, given $\epsilon > 0$ , I need to find $\delta$ such that:

$|a^x - 0| < \epsilon$ whenever $x < \delta$

And here's where I am stuck. I think I'll have to take a log in there somewhere. Wish my textbook wasn't so scarce on examples (Stewart's Calculus 4th edition). I sort of get limits where x->c, but I'm totally lost on the ones with infinities, and the differences between handling + and - infinities. Can't find any examples online of ones as x-> -infinity, which surely might help. Nor are the two videos at Khan Academy useful in this case.

I bet you guys get a lot of these. Surprisingly confusing for what seems a relatively simple concept, at first. Help, hints, pointers, nudges and taunts all appreciated.

**Ackbeet** · December 23rd 2010, 04:40 PM

If you think of $\delta$ as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that $a>0,$ right? Otherwise, you have complex numbers floating around.

How can you simplify $|a^{x}-0|<\epsilon?$

If you can find a $\delta=\delta(\epsilon)$ that works, you'll be done. How could you do that?

**Grep** · December 23rd 2010, 05:18 PM

Originally Posted by Ackbeet

If you think of $\delta$ as a large negative number, and getting more and more negative, you'll be on the right track; also, you have to assume that $a>0,$ right? Otherwise, you have complex numbers floating around.

How can you simplify $|a^{x}-0|<\epsilon?$

Well, first and obviously, $|a^{x} - 0| < \epsilon \Rightarrow |a^{x}| < \epsilon$ .

Then, I can assume that $a^{x} > 0$ so I have just:

$a^{x} < \epsilon$ whenever $x < \epsilon$ .

Originally Posted by Ackbeet

If you can find a $\delta=\delta(\epsilon)$ that works, you'll be done. How could you do that?

We want:

$a^{x} < \epsilon$ whenever $x < \epsilon$

Or:

$log_{a}(a^{x}) = log_{a}(\epsilon) \Rightarrow x = log_{a}(\epsilon)$

So I should try $\delta = log_{a}(\epsilon)$ .

Given $\epsilon > 0$ , we choose $\delta = log_{a}(\epsilon)$ . Let $x < \epsilon$ . Then

$|a^{x} - 0| = a^{x} < a^{\delta} = a^{log_{a}(\epsilon)} = \epsilon$

Thus

$|a^{x} - 0| < \epsilon$

Is that right? Egads, I think I just did it. Hey, I think I smell burnt toast!

**chisigma** · December 23rd 2010, 07:32 PM

Originally Posted by Grep

I think Epsilon-delta proofs and I should spend some quality time together.

The problem is I need to prove that:

$\lim_{x \to -\infty} a^x = 0$

...

... that's true if $a >1$ ... of course...

Merry Christmas from Italy

$\chi$ $\sigma$

**Grep** · December 23rd 2010, 07:45 PM

Originally Posted by chisigma

... that's true if $a >1$ ... of course...

Good point, the problem description says "Let a > 1", which I should have stated. Apologies for leaving out an important part of the problem description.

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